Finding rank of linear tranformation without a matrix? A practice problem I have asks us to find the rank of the linear transformation:
$$T(x, y, z) = (x - 2y + z, 2x + y + z)$$
What I'm confused on is, thus far in my class, as far as I can recall, we've only ever found the rank of linear transformations in the form $T(x) = Ax$, where you have a matrix $--$ and you put the matrix in reduced row-echelon form and see how many leading $1$'s there are in order to determine the rank. So since we don't have a matrix here, I'm not sure how I would solve this.
I'm getting the sense that it's possible to put ANY linear transformation into matrix form, but the problem is my book hasn't reached that point yet; I'm pretty sure that's in the next section. So I don't know how to represent this transformation with a matrix (and I don't think I'm supposed to since we haven't learned that yet), but I don't know how I would solve this problem without doing that.
The problem also says NOT to compute the kernel, so I can't find the kernel and then use the rank-nullity theorem to find the rank, either.
I'm at all loss as to how I would do this. I've flipped through the section and all the problems in my book and I don't see any examples that compute the rank WITHOUT having a matrix.
Is there another way to compute the rank of this that I'm missing?
 A: Quite simple: $T(-1,0,2)=(1,0)$ and $T(1,0,-1)=(0,1)$, i.e. the canonical basis of $\mathbf R^2$ is attained, so the rank is $2$.
A: Vectors $\begin{pmatrix}1\\2\end{pmatrix}$ and $\begin{pmatrix}-2\\1\end{pmatrix}$,  being "the images of something", more precisely of : $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ resp. are in the image space.
Moreover, these vectors are independent. Thus, the rank r, which is by definition the dimension of the image space, is at least 2 but cannot be more than that because it cannot exceed the dimension of the "ambient" space $\mathbb{R^2}.$ 
A confirmation is brought by the fact that $\begin{pmatrix}-3\\1\\5\end{pmatrix}$ belongs to the kernel. Thus the dimension of the kernel is $k \geq 1$ and in fact no more than $1$ because of the rank-nullity theorem ($r+k=2+1=3$.)

Edit: I place here the matrix representation of the linear transformation:
$$\begin{pmatrix}1&-2&1\\2&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$
A: $(x,y,z) \in ker(T)$ iff $(x - 2y + z, 2x + y + z)=(0,0)$
Its your turn to solve the linear system
$x - 2y + z=0 $
$2x + y + z=0$.
Then you should see that $\dim ker(T)=1$. From 
$3= \dim \mathbb R^3= \dim ker(T)+rank(T)$ we get $rank(T)=2$.
A: The rank is either $0, 1$ or $2$. $\begin{cases} \operatorname{rank}(T) = 0\Rightarrow T = (0,0)\\ \operatorname{rank}(T) = 1 \Rightarrow x - 2y + z = \alpha(2x + y + z) & \text {not possible} \\ \operatorname{rank}(T) = 2 &\text{the only possibility left}\end{cases}$
